The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg, respectivelyat 350 K. Calculate the composition of liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition of the mixture in vapour phase.

The vapour pressures of pure liquids A and B are 450 mm and 700 mm of

1.	The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg, respectivelyat 350 K. Calculate the composition of liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition of the mixture in vapour phase.
Answer» Solution : Let the mole fraction of A be `x_A`and that of B be `x_B` . Let the vapour pressure of component `A = p_A^0` Vapour pressure of component `B = p_B^0` Total pressure` = P_("total")` `P_("total") = x_Ap_A^0 + x_Bp_B^0 = (1-x_B)p_A^0 + x_B p_B^0` `p_("total") = p_A^0 + (p_B^0 + p_A^0) x_B` Substituting the values, we have `600 = 450 + (700 - 450)x_B` `250x_B = 150 " or " x_B = 0.6` THUS, a mixture of A and B with B having mole fraction of 0.6 will have the total vapour pressure of 600 MM. Composition of the mixture in vapour phase Use the relation : `p = yp_("total")`, where y is the mole fraction in vapour phase `p_A = 0.4 xx 450 = 180 mm` `p_B = 0.6 xx 700 = 420 mm` `y_A = (p_A)/(p_("total")) = 180/600 = 0.3` `y_B = (p_B)/(p_("total")) = 420/600 = 0.7`